Title MSTE 35 Solutions.pdf ✅

35 Surveying: Horizontal Curves 2 Solutions SITUATION 1. The perpendicular distance between two parallel tangents of the reserve curve is 35 m. The azimuth of the back tangent is 270° while the common tangent is 300°. The first radius of the curve is 160 m and the stationing of PRC is 2+578. Determine the following: ▣ 1. Radius of the second curve. [SOLUTION] T1 + T2 = 35 m sin 30° T1 = R1 tan I1 2 = 160 tan 30° 2 160 tan 30° 2 + R2 tan 30° 2 = 35 m sin 30° R2 = 101.2436 m ▣ 2. Stationing of PC. [SOLUTION] Sta PC = Sta PRC − L1 Sta PC = (2 + 578) − π(160 m)(30°) 180° = 2 + 494.224 ▣ 3. Stationing of PT. [SOLUTION] Sta PT = Sta PRC + L2 Sta PT = (2 + 578) + π(101.2436 m)(30°) 180° = 2 + 631.011
SITUATION 2. A spiral curve having a length of 100 m is to be laid out in a certain portion of road. The degree of the central curve is 6°. ▣ 4. Find the offset distance at the first quarter point of spiral. [SOLUTION] Rc = 3600 π(6) = 190.986 m x = L 3 6RcLc x = ( 100 m 4 ) 3 6(190.986 m)(100 m) = 0.136 m ▣ 5. Determine the spiral angle at the third quarter point of the spiral. [SOLUTION] θ = L 2 2RcLc θ = ( 3 4 × 100 m) 2 2(190.986 m)(100 m) θ = 3π 64 rad ≈ 8.4375° ▣ 6. Compute the maximum speed of the car that could pass through the spiral without skidding. [SOLUTION] L = 0.036 v 3 R 100 m = 0.036 v 3 190.986 v = 80.953 kph SITUATION 3. A spiral easement curve has a spiral angle at SC of 12° and an offset distance at SC equal to 3.4 m. Distance along tangent up to SC is 79.62 m.
θ = 12°, x = 3.4 m, y = 79.62 m ▣ 7. What is the length of the short tangent? [SOLUTION] ST = x sin θ ST = 3.4 m sin 12° = 16.353 m ▣ 8. What is the length of the long tangent? [SOLUTION] LT = y − x tan θ LT = 79.62 m − 3.4 m tan 12° LT = 63.62 m ▣ 9. What is the length of throw? [SOLUTION] p = xc 4 p = 3.4 m 4 = 0.85 m